Finding all values $p$ for which $int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dx$ converges
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$begingroup$
I've been stuck for a while with this exercise. Find all positive real values $p$ for which the integral
$$int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dx$$
converges. So far I've came up with this:
$$ int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dxgeint_e^{+infty} frac{1}{(1+x^3)^frac{1}{p}}dx
\
text{Take the limit of dividing the second function by }frac{1}{x^frac{3}{p}}
\
lim_{xto{+infty}}frac{x^frac{3}{p}}{(1+x^3)^frac{1}{p}}=1
\
int_e^{+infty}frac{1}{x^frac{3}{p}}text{ Diverges } leftrightarrow pge3
$$
So, when $pge3$ my p-series diverges, which means my lower boundary for the main function diverges, which implies that the main function diverges. But I'm unable to prove any other implication. Any suggestions?
calculus integration convergence improper-integrals
edited Nov 30 '18 at 5:40
Olivier Oloa
108k17177294
asked Nov 30 '18 at 0:20
twkmztwkmz
545
$endgroup$
add a comment |
$begingroup$
I've been stuck for a while with this exercise. Find all positive real values $p$ for which the integral
$$int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dx$$
converges. So far I've came up with this:
$$ int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dxgeint_e^{+infty} frac{1}{(1+x^3)^frac{1}{p}}dx
\
text{Take the limit of dividing the second function by }frac{1}{x^frac{3}{p}}
\
lim_{xto{+infty}}frac{x^frac{3}{p}}{(1+x^3)^frac{1}{p}}=1
\
int_e^{+infty}frac{1}{x^frac{3}{p}}text{ Diverges } leftrightarrow pge3
$$
So, when $pge3$ my p-series diverges, which means my lower boundary for the main function diverges, which implies that the main function diverges. But I'm unable to prove any other implication. Any suggestions?
calculus integration convergence improper-integrals
edited Nov 30 '18 at 5:40
Olivier Oloa
108k17177294
asked Nov 30 '18 at 0:20
twkmztwkmz
545
$endgroup$
add a comment |
$begingroup$
I've been stuck for a while with this exercise. Find all positive real values $p$ for which the integral
$$int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dx$$
converges. So far I've came up with this:
$$ int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dxgeint_e^{+infty} frac{1}{(1+x^3)^frac{1}{p}}dx
\
text{Take the limit of dividing the second function by }frac{1}{x^frac{3}{p}}
\
lim_{xto{+infty}}frac{x^frac{3}{p}}{(1+x^3)^frac{1}{p}}=1
\
int_e^{+infty}frac{1}{x^frac{3}{p}}text{ Diverges } leftrightarrow pge3
$$
So, when $pge3$ my p-series diverges, which means my lower boundary for the main function diverges, which implies that the main function diverges. But I'm unable to prove any other implication. Any suggestions?
calculus integration convergence improper-integrals
edited Nov 30 '18 at 5:40
Olivier Oloa
108k17177294
asked Nov 30 '18 at 0:20
twkmztwkmz
545
$endgroup$
I've been stuck for a while with this exercise. Find all positive real values $p$ for which the integral
$$int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dx$$
converges. So far I've came up with this:
$$ int_e^{+infty} frac{ln(x)}{(1+x^3)^frac{1}{p}}dxgeint_e^{+infty} frac{1}{(1+x^3)^frac{1}{p}}dx
\
text{Take the limit of dividing the second function by }frac{1}{x^frac{3}{p}}
\
lim_{xto{+infty}}frac{x^frac{3}{p}}{(1+x^3)^frac{1}{p}}=1
\
int_e^{+infty}frac{1}{x^frac{3}{p}}text{ Diverges } leftrightarrow pge3
$$
So, when $pge3$ my p-series diverges, which means my lower boundary for the main function diverges, which implies that the main function diverges. But I'm unable to prove any other implication. Any suggestions?
calculus integration convergence improper-integrals
calculus integration convergence improper-integrals
edited Nov 30 '18 at 5:40
Olivier Oloa
108k17177294
asked Nov 30 '18 at 0:20
twkmztwkmz
545
edited Nov 30 '18 at 5:40
Olivier Oloa
108k17177294
asked Nov 30 '18 at 0:20
twkmztwkmz
545
edited Nov 30 '18 at 5:40
Olivier Oloa
108k17177294
edited Nov 30 '18 at 5:40
Olivier Oloa
108k17177294
edited Nov 30 '18 at 5:40
Olivier Oloa
108k17177294
108k17177294
asked Nov 30 '18 at 0:20
twkmztwkmz
545
asked Nov 30 '18 at 0:20
twkmztwkmz
545
asked Nov 30 '18 at 0:20
twkmztwkmz
545
545
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint. One may use that (with an integration by parts)
$$
int_e^infty frac{ln x}{x^a},dx qquad text{converges iff} qquad a>1.
$$ then one may observe that, as $x to infty$,
$$
frac{ln x}{(1+x^3)^frac{1}{p}} sim frac{ln x}{x^{3/p}}.
$$
answered Nov 30 '18 at 0:26
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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votes
$begingroup$
Hint. One may use that (with an integration by parts)
$$
int_e^infty frac{ln x}{x^a},dx qquad text{converges iff} qquad a>1.
$$ then one may observe that, as $x to infty$,
$$
frac{ln x}{(1+x^3)^frac{1}{p}} sim frac{ln x}{x^{3/p}}.
$$
answered Nov 30 '18 at 0:26
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
add a comment |
$begingroup$
Hint. One may use that (with an integration by parts)
$$
int_e^infty frac{ln x}{x^a},dx qquad text{converges iff} qquad a>1.
$$ then one may observe that, as $x to infty$,
$$
frac{ln x}{(1+x^3)^frac{1}{p}} sim frac{ln x}{x^{3/p}}.
$$
answered Nov 30 '18 at 0:26
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
add a comment |
$begingroup$
Hint. One may use that (with an integration by parts)
$$
int_e^infty frac{ln x}{x^a},dx qquad text{converges iff} qquad a>1.
$$ then one may observe that, as $x to infty$,
$$
frac{ln x}{(1+x^3)^frac{1}{p}} sim frac{ln x}{x^{3/p}}.
$$
answered Nov 30 '18 at 0:26
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
Hint. One may use that (with an integration by parts)
$$
int_e^infty frac{ln x}{x^a},dx qquad text{converges iff} qquad a>1.
$$ then one may observe that, as $x to infty$,
$$
frac{ln x}{(1+x^3)^frac{1}{p}} sim frac{ln x}{x^{3/p}}.
$$
answered Nov 30 '18 at 0:26
Olivier OloaOlivier Oloa
108k17177294
answered Nov 30 '18 at 0:26
Olivier OloaOlivier Oloa
108k17177294
answered Nov 30 '18 at 0:26
Olivier OloaOlivier Oloa
108k17177294
answered Nov 30 '18 at 0:26
Olivier OloaOlivier Oloa
108k17177294
108k17177294
add a comment |
add a comment |
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